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Long Division of Improper Rational Functions

We are about to look at a very important technique of integration known as Integration with Partial Fractions, however, we will first look at a not-so-calculus technique often necessary to apply the integration with partial fractions method. We begin with a few important definitions.

Definition: If $f$ is a rational function, then the Degree of $f$ often denoted $\mathrm{deg} (f)$ is equal to the highest value exponent in $f$.

Definition: If $f(x) = \frac{P(x)}{Q(x)}$ is a rational function, and $\mathrm{deg} (P) ≥ \mathrm{deg} (Q)$, then we say that $f$ is an Improper Rational Function. Furthermore, if $\mathrm{deg} (P) < \mathrm{deg} (Q)$, then we say that $f$ is a Proper Rational Function.

The technique of long division will only apply to improper rational functions as we will see.

Note: In our example, the degrees of each term in $P(x)$ and $Q(x)$ are descending. For this algorithm to work properly, it is best to reorder the terms of $P(x)$ and $Q(x)$ if they're not descending in degree.

Take the first term of $P$ and divide it by the first term of $Q(x)$. In our case, we get that $x^3$ divided by $x$ is $x^2$. Now multiply $x^2$ by the divisor $Q(x)$ and find the difference $P(x) - x^2Q(x)$ as follows:

Now take the first term of this new function, in our case, it is $3x^2$, and divide it by the first term in $Q(x)$, which of course is still $x$. We note that $3x^2$ divided by $x$ equals $3x$. Now take $[P(x) - x^2Q(x)]$ and subtract $3xQ(x)$ as follows:

Now take the first term of this new function, in our case, it is $3x$, and divide it by the first term in $Q(x)$, which is $x$. We note that $3x$ divided by $x$ is $3$. Now take $[[P(x) - x^2Q(x)] - 3xQ(x)] - 3Q(x)$ as follows: